# Tangent line of curve of intersection

AvalonX

## Homework Statement

Find the parametric equations of the line tangent to the curve of intersection of the paraboloid
z = x² + y² and the ellipsoid 4x² + y² + z² = 9 at the point ( -1, 1, 2 ).

## Homework Equations

Probable use of the gradient vector (as this is the chapter we are in)

## The Attempt at a Solution

I have found what I believe to be the curve of intersection by setting the equations of the surfaces equal to each other:

x² + y² - z = 4x² + y² + z² - 9
9 = 3x² + z² + z
F(x,y,z) = 3x² + z² + z - 9 (as a function)

And the corresponding gradient vector:
< 6x, 0, 2z + 1 >

But I cannot figure out how to convert this from cartesian to polar (or if I even need to) and get the correct answer. I've tried several paths (i.e. substituting x=rcosσ) but my answer never matches the expected answer, which contains no trigonometry. Does the correct approach involve substituting x² + y² for z? This seems to be the most logical approach but I haven't been able to apply it correctly.

## Answers and Replies

Gold Member
It is unclear to me why you want the tangent line in polar coordinates. The text seems only to call for a parametric line, eg. $\bar{r} = \bar{k} t +\bar{r}_0$.

AvalonX
Your right! I was thinking of a vector equation. Then I suppose the parametric equations would look something like: x = -1 +3t², y = 1, z = 3 + t² ? But that still doesn't contain the expected y component, I must be doing something wrong yet.

Gold Member
A line parametric in, say, t should be linear in t, otherwise its not a line. Think about which point the line passes through and what vector direction it "follows" from that point as t varies.

AvalonX
I remember why I wanted the vector equation. The only way I can think to find the tangent line is by taking the derivative of a vector equation.

AvalonX
Never mind. By crossing the gradients I found the tangent vector. This makes perfect sense, it was so simple :D :D